Problem: Solve for $x$ : $2x^2 - 6x - 36 = 0$
Explanation: Dividing both sides by $2$ gives: $ x^2 {-3}x {-18} = 0 $ The coefficient on the $x$ term is $-3$ and the constant term is $-18$ , so we need to find two numbers that add up to $-3$ and multiply to $-18$ The two numbers $-6$ and $3$ satisfy both conditions: $ {-6} + {3} = {-3} $ $ {-6} \times {3} = {-18} $ $(x {-6}) (x + {3}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -6) (x + 3) = 0$ $x - 6 = 0$ or $x + 3 = 0$ Thus, $x = 6$ and $x = -3$ are the solutions.